关于三角函数的等式证明求证:三角形ABC中,tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(A/2)·tan(C/2)=1
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![关于三角函数的等式证明求证:三角形ABC中,tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(A/2)·tan(C/2)=1](/uploads/image/z/5912764-52-4.jpg?t=%E5%85%B3%E4%BA%8E%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0%E7%9A%84%E7%AD%89%E5%BC%8F%E8%AF%81%E6%98%8E%E6%B1%82%E8%AF%81%EF%BC%9A%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2Ctan%28A%2F2%29%C2%B7tan%28B%2F2%29%2Btan%28B%2F2%29%C2%B7tan%28C%2F2%29%2Btan%28A%2F2%29%C2%B7tan%28C%2F2%29%3D1)
关于三角函数的等式证明求证:三角形ABC中,tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(A/2)·tan(C/2)=1
关于三角函数的等式证明
求证:三角形ABC中,
tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(A/2)·tan(C/2)=1
关于三角函数的等式证明求证:三角形ABC中,tan(A/2)·tan(B/2)+tan(B/2)·tan(C/2)+tan(A/2)·tan(C/2)=1
因为在△ABC中,A+B+C=180°
所以:(A+B+C)/2=90°
所以,(A/2)=90°-(B+C)/2
那么:
tan(A/2)=tan[90°-(B+C)/2]=cot[(B+C)/2]=1/tan[(B+C)/2]
=1/{[tan(B/2)+tan(C/2)]/[1-tan(B/2)tan(C/2)]}
=[1-tan(B/2)tan(C/2)]/[tan(B/2)+tan(C/2)]……………(1)
上述等式左边
tan(A/2)*tan( B/2)+tan(B/2)tan(C/2)+tan(C/2)*tan(A/2)
=tan(A/2)*[tan(B/2)+tan(C/2)]+tan(B/2)tan(C/2)
将(1)式代入上式,则:
=[1-tan(B/2)tan(C/2)]+tan(B/2)tan(C/2)
=1
=右边
所以,命题成立
tan(b/2)tan(c/2)+tan(c/2)tan(a/2)
=tan(c/2)[tan(a/2)+tan(b/2)]
=tan[90-(a+b)/2]×[tan(a/2)+tan(b/2)]
=cot[(a+b)/2]×[tan(a/2)+tan(b/2)]
=[tan(a/2)+tan(b/2)]/tan(a/2+b/2)
=1-tan(...
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tan(b/2)tan(c/2)+tan(c/2)tan(a/2)
=tan(c/2)[tan(a/2)+tan(b/2)]
=tan[90-(a+b)/2]×[tan(a/2)+tan(b/2)]
=cot[(a+b)/2]×[tan(a/2)+tan(b/2)]
=[tan(a/2)+tan(b/2)]/tan(a/2+b/2)
=1-tan(a/2)tan(b/2)
∴tan(a/2)tan(b/2)+tan(b/2)tan(c/2)+tan(c/2)tan(a/2) = tan(a/2)tan(b/2)+1-tan(a/2)tan(b/2) = 1
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